Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $x \neq 0$. $n = \dfrac{x^2 - 100}{3x^2 + 30x} \times \dfrac{x - 5}{x - 10} $
Answer: First factor the quadratic. $n = \dfrac{(x - 10)(x + 10)}{3x^2 + 30x} \times \dfrac{x - 5}{x - 10} $ Then factor out any other terms. $n = \dfrac{(x - 10)(x + 10)}{3x(x + 10)} \times \dfrac{x - 5}{x - 10} $ Then multiply the two numerators and multiply the two denominators. $n = \dfrac{ (x - 10)(x + 10) \times (x - 5) } { 3x(x + 10) \times (x - 10) } $ $n = \dfrac{ (x - 10)(x + 10)(x - 5)}{ 3x(x + 10)(x - 10)} $ Notice that $(x + 10)$ and $(x - 10)$ appear in both the numerator and denominator so we can cancel them. $n = \dfrac{ \cancel{(x - 10)}(x + 10)(x - 5)}{ 3x(x + 10)\cancel{(x - 10)}} $ We are dividing by $x - 10$ , so $x - 10 \neq 0$ Therefore, $x \neq 10$ $n = \dfrac{ \cancel{(x - 10)}\cancel{(x + 10)}(x - 5)}{ 3x\cancel{(x + 10)}\cancel{(x - 10)}} $ We are dividing by $x + 10$ , so $x + 10 \neq 0$ Therefore, $x \neq -10$ $n = \dfrac{x - 5}{3x} ; \space x \neq 10 ; \space x \neq -10 $